Question

A sodium ion, Na+, with a charge of 1.6×10−19C and a chloride ion, Cl− , with charge of −1.6×10−19C, are separated by a distance of 1.1nm. How much work would be required to increase the separation of the two ions to an infinite distance?

Answer 1

Answer:

$W\geq 2.1x10^{-19}J$

Explanation:

Due to Coulomb´s law electric force can be described by the formula $F=K\frac{q_{1}.q_{2}}{r^{2}}$, where K is the Coulomb´s constant ($9x10^{9} N\frac{m^{2} }{C^{2} }$), $q_{1}$= Charge 1 (Na+ in this case), $q_{2}$ is the charge 2 (Cl-) and r is the distance between both charges.

Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is $W=W_{f} -W_{i}$.

so we have $W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]$

Given that ri= 1.1nm= $1.1x10^{-9}m$ and rf= infinite distance

$W=(9x10^{9})(1.6x10^{-19})(-1.6x10^{-19})[\frac{1}{\alpha }-\frac{1}{(1.1x10^{-9})}]=2.1x10^{-19}J$